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class="site-page" href="javascript:void(0);"><i class="fa-fw fa fa-list"></i><span> 清单</span><i class="fas fa-chevron-down expand"></i></a><ul class="menus_item_child"><li><a class="site-page child" href="/music/"><i class="fa-fw fa fa-music"></i><span> 音乐</span></a></li><li><a class="site-page child" href="/movies/"><i class="fa-fw fa fa-film"></i><span> 电影</span></a></li><li><a class="site-page child" href="/Gallery/"><i class="fa-fw fas fa-images"></i><span> 照片</span></a></li></ul></div></div><div id="toggle-menu"><a class="site-page"><i class="fas fa-bars fa-fw"></i></a></div></div></nav><div id="post-info"><h1 class="post-title">汉诺塔非递归实现 C语言版</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2020-01-20T16:00:00.000Z" title="发表于 2020-01-21 00:00:00">2020-01-21</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2020-02-10T03:01:22.996Z" title="更新于 2020-02-10 11:01:22">2020-02-10</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E7%AE%97%E6%B3%95%E9%A2%98/">算法题</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-wordcount"><i class="far fa-file-word fa-fw post-meta-icon"></i><span class="post-meta-label">字数总计:</span><span class="word-count">788</span><span class="post-meta-separator">|</span><i class="far fa-clock fa-fw post-meta-icon"></i><span class="post-meta-label">阅读时长:</span><span>3分钟</span></span><span class="post-meta-separator">|</span><span class="post-meta-pv-cv" id="" data-flag-title="汉诺塔非递归实现 C语言版"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="汉诺塔非递归实现-C语言版"><a href="#汉诺塔非递归实现-C语言版" class="headerlink" title="汉诺塔非递归实现 C语言版"></a><strong><em>汉诺塔非递归实现 C语言版</em></strong></h1><p>我上一篇博客是汉诺塔C语言递归实现，非递归和递归想法一样。这里不再赘述，直接链接转到：</p>
<p><a target="_blank" rel="noopener" href="https://blog.csdn.net/VistorsYan/article/details/102765478">汉诺塔递归实现 C语言版</a></p>
<p>   递归实现固然好理解，但是n的值越大，空间和时间上都是极大的消耗，最终可能导致程序直接崩溃。<br>在以后的做题或者是面试中，不推荐用递归方法做，所以要写出对应的非递归方法。</p>
<p>  某次上课无意间听到老师说了这样一句话：任何递归法都可以用循环的方法进行非递归实现，然后回头找了找汉诺塔非递归的资料，整理整理，搞出了一个c实现的非递归方法</p>
<figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">#include&lt;stdio.h&gt;</span></span><br><span class="line"><span class="comment">#include &lt;stdlib.h&gt;</span></span><br><span class="line"><span class="comment">#define MaxSize 100</span></span><br><span class="line">typedef struct&#123;</span><br><span class="line">     int N;</span><br><span class="line">     char A;        //起始柱</span><br><span class="line">     char B;        //借助柱</span><br><span class="line">     char C;        //目标柱</span><br><span class="line">&#125;ElementType;</span><br><span class="line">typedef struct &#123;</span><br><span class="line">    ElementType Data[MaxSize];</span><br><span class="line">    int top;</span><br><span class="line">&#125;Stack;//汉诺塔问题的结构类型</span><br><span class="line">void Push(Stack *PtrS, ElementType item)&#123;</span><br><span class="line">     //入栈操作</span><br><span class="line">     <span class="keyword">if</span> (PtrS-&gt;top == MaxSize)</span><br><span class="line">     &#123;</span><br><span class="line">         <span class="built_in">printf</span>(<span class="string">&quot;The stack is full!\n&quot;</span>);</span><br><span class="line">         <span class="built_in">return</span>;</span><br><span class="line">     &#125;</span><br><span class="line">     <span class="keyword">else</span></span><br><span class="line">     &#123;</span><br><span class="line">         PtrS-&gt;Data[++(PtrS-&gt;top)] = item;</span><br><span class="line">         <span class="built_in">return</span>;</span><br><span class="line">     &#125;</span><br><span class="line">&#125;</span><br><span class="line">ElementType Pop(Stack *PtrS)&#123;</span><br><span class="line">    <span class="keyword">if</span> (PtrS-&gt;top == -1)</span><br><span class="line">      &#123;</span><br><span class="line">          <span class="built_in">printf</span>(<span class="string">&quot;The stack is empty!\n&quot;</span>);</span><br><span class="line">          <span class="built_in">exit</span>(1);   //直接终止程序，一般不会出现这个错误</span><br><span class="line">      &#125;</span><br><span class="line">      <span class="keyword">else</span></span><br><span class="line">      &#123;</span><br><span class="line">          PtrS-&gt;top--;</span><br><span class="line">         <span class="built_in">return</span> (PtrS-&gt;Data[PtrS-&gt;top + 1]);        //或者是<span class="built_in">return</span> PtrS-&gt;Data[PtrS-&gt;top--];</span><br><span class="line">      &#125;</span><br><span class="line">&#125;</span><br><span class="line">//借助栈的非递归实现</span><br><span class="line"> void Hanoi(int n)&#123;</span><br><span class="line">    ElementType P, toPush;</span><br><span class="line">    Stack S;</span><br><span class="line"></span><br><span class="line">    P.N = n; P.A = <span class="string">&#x27;a&#x27;</span>; P.B = <span class="string">&#x27;b&#x27;</span>; P.C = <span class="string">&#x27;c&#x27;</span>;</span><br><span class="line">    S.top = -1;</span><br><span class="line"></span><br><span class="line">     Push(&amp;S, P);</span><br><span class="line">     <span class="keyword">while</span> (S.top != -1)        //当堆栈不为空时</span><br><span class="line">     &#123;</span><br><span class="line">         P = Pop(&amp;S);//出栈</span><br><span class="line">         <span class="keyword">if</span> (P.N == 1)//当只剩一个盘子时，直接由当前柱移动到目的柱</span><br><span class="line">             <span class="built_in">printf</span>(<span class="string">&quot;%c -&gt; %c\n&quot;</span>, P.A, P.C);</span><br><span class="line">         <span class="keyword">else</span></span><br><span class="line">         &#123;</span><br><span class="line">             toPush.N = P.N - 1;</span><br><span class="line">             toPush.A = P.B; toPush.B = P.A; toPush.C = P.C;</span><br><span class="line">             Push(&amp;S, toPush);        //将第三步(n - 1, b, a, c)入栈</span><br><span class="line">             toPush.N = 1;</span><br><span class="line">             toPush.A = P.A; toPush.B = P.B; toPush.C = P.C;</span><br><span class="line">             Push(&amp;S, toPush);        //将第二步1, a, b, c)入栈</span><br><span class="line">             toPush.N = P.N - 1;</span><br><span class="line">             toPush.A = P.A; toPush.B = P.C; toPush.C = P.B;</span><br><span class="line">             Push(&amp;S, toPush);        //将第一步(n - 1, a, c, b)入栈</span><br><span class="line">         &#125;</span><br><span class="line">     &#125;</span><br><span class="line"> &#125;</span><br><span class="line">int <span class="function"><span class="title">main</span></span>()&#123;</span><br><span class="line">    int n;</span><br><span class="line">    scanf(<span class="string">&quot;%d&quot;</span>, &amp;n);</span><br><span class="line">    <span class="keyword">if</span> (n &lt;= 0)<span class="built_in">return</span> 0;</span><br><span class="line">    <span class="keyword">else</span> Hanoi(n);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>还是三个步骤：<br>1.将n-1个盘子由a柱借助c柱移动到b柱<br>2.将最下面的盘子由a柱直接移动到c柱<br>3.将那n-1个盘子在由b柱借助a柱移动到c柱</p>
<p>因为这个是出栈时的操作，所以入栈时要到着写</p>
<h2 id="简要解释一下（因为跟递归思路差不多）"><a href="#简要解释一下（因为跟递归思路差不多）" class="headerlink" title="简要解释一下（因为跟递归思路差不多）"></a>简要解释一下（因为跟递归思路差不多）</h2><p>如果n不等于一时，就意味着，以上的n-1个盘子，都要做上述所说的三个步骤，知道n等于1时，直接移动到目的柱。<br>因此，移动次数最多的是最上边的那个盘子，移动次数最少的是最下面的那个盘子，只需要移动一次</p>
<p>利用结构体数组更便于理解。</p>
<p>本文为原创，如有问题欢迎评论区留言。</p>
</article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href="mailto:undefined">Charles Yan</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://www.charlesyan.cn/2020/01/21/%E6%B1%89%E8%AF%BA%E5%A1%94%E9%9D%9E%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0c%E8%AF%AD%E8%A8%80%E7%89%88/">https://www.charlesyan.cn/2020/01/21/%E6%B1%89%E8%AF%BA%E5%A1%94%E9%9D%9E%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0c%E8%AF%AD%E8%A8%80%E7%89%88/</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://www.charlesyan.cn" target="_blank">路漫漫其修远兮，吾将上下而求索</a>！</span></div></div><div class="tag_share"><div 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